calculate-0-t-2-e-t-1-dt-interms-of-3-

Question Number 53783 by maxmathsup by imad last updated on 25/Jan/19

c a l c u l a t e \int_{0}^{\infty} \frac{t^{2}}{e^{t} - 1} d t i n t e r m s o f ξ (3)

Answered by Smail last updated on 26/Jan/19

A \int_{0}^{\infty} \frac{t^{2}}{e^{t} - 1} d t = \int_{0}^{\infty} \frac{t^{2} e^{- t}}{1 - e^{- t}} d t

= \int_{0}^{\infty} t^{2} e^{- t} \underset{n = 0}{\sum^{\infty}} e^{- n t} d t

= \underset{n = 0}{\sum^{\infty}} \int_{0}^{\infty} t^{2} e^{- (n + 1) t} d t

b y p a r t s

u = t^{2} \Rightarrow u^{'} = 2 t

v^{'} = e^{- (n + 1) t} \Rightarrow v = \frac{- 1}{n + 1} e^{- (n + 1) t}

A = \underset{n = 0}{\sum^{\infty}} \frac{2}{n + 1} \int_{0}^{\infty} {t e}^{- (n + 1) t} d t w i t h ({[t^{2} e^{- (n + 1) t}]}_{0}^{\infty} = 0)

b y p a r t s

A = \underset{n = 0}{\sum^{\infty}} \frac{2}{{(n + 1)}^{2}} \int_{0}^{\infty} e^{- (n + 1) t} d t

= \underset{n = 0}{\sum^{\infty}} \frac{2}{{(n + 1)}^{3}} {[e^{- (n + 1) t}]}_{0}^{\infty} = 2 \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + 1)}^{3}}

= 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} = 2 ξ (3)

Commented by maxmathsup by imad last updated on 26/Jan/19

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